Gina wrote the following paragraph to prove that the segment joining the midpoints of two sides of a triangle is parallel to the third side: given: δabc prove: the midsegment between sides line segment ab and line segment bc is parallel to side line segment ac. draw δabc on the coordinate plane with point a at the origin (0, 0). let point b have the ordered pair (x1, y1) and locate point c on the x-axis at (x2, 0). point d is the midpoint of line segment ab with coordinates at ordered pair the quantity 0 plus x sub 1, divided by 2. the quantity 0 plus y sub 1, divided by 2 by the slope formula. point e is the midpoint of line segment bc with an ordered pair of ordered pair the quantity of x sub 1 plus x sub 2 divided by 2. the quantity of 0 plus y sub 1 divided by 2 by the slope formula. the slope of line segment de is found to be 0 through the application of the slope formula: the difference of y sub 2 and y sub 1, divided by the difference of x sub 2 and x sub 1 is equal to the difference of the quantity 0 plus y sub 1, divided by 2, and the quantity 0 plus y sub 1, divided by 2, divided by the difference of the quantity x sub 1 plus x sub 2, divided by 2 and the quantity 0 plus x sub 1, divided by 2 is equal to 0 divided by the quantity x sub 2 divided by 2 is equal to 0 when the slope formula is applied to line segment ac the difference between y sub 2 and y sub 1, divided by the difference of x sub 2 and x sub 1 is equal to the difference of 0 and 0, divided by the difference of x sub 2 and 0 is equal to 0 divided by x sub 2 is equal to 0, its slope is also 0. since the slope of line segment de and line segment ac are identical, line segment de and line segment ac are parallel by the definition of parallel lines. which statement corrects the flaw in gina's proof?

when functions are first introduced, you will probably have some simplistic "functions" and relations to deal with, usually being just sets of points. these won't be terribly useful or interesting functions and relations, but your text wants you to get the idea of what the domain and range of a function are. small sets of points are generally the simplest sorts of relations, so your book starts with those.

step-by-step explanation:

state the domain and range of the following relation. is the relation a function?

{(2, –3), (4, 6), (3, –1), (6, 6), (2, 3)}

the above list of points, being a relationship between certain x's and certain y's, is a relation. the domain is all the x-values, and the range is all the y-values. to give the domain and the range, i just list the values without duplication:

domain: {2, 3, 4, 6}

range: {–3, –1, 3, 6}

(it is customary to list these values in numerical order, but it is not required. sets are called "unordered lists", so you can list the numbers in any order you feel like. just don't duplicate: technically, repetitions are okay in sets, but most instructors would count off for this.)

while the given set does indeed represent a relation (because x's and y's are being related to each other), the set they gave me contains two points with the same x-value: (2, –3) and (2, 3). since x = 2 gives me two possible destinations (that is, two possible y-values), then this relation is not a function.

note that all i had to do to check whether the relation was a function was to look for duplicate x-values. if you find any duplicate x-values, then the different y-values mean that you do not have a function. remember: for a relation to be a function, each x-value has to go to one, and only one, y-value.

state the domain and range of the following relation. is the relation a function?

{(–3, 5), (–2, 5), (–1, 5), (0, 5), (1, 5), (2, 5)}

i'll just list the x-values for the domain and the y-values for the range:

1 + (6 / x - 1)

step-by-step explanation:

divide using polynomial division.